## Calculus (3rd Edition)

$\ln 2$
Let $u=\ln t$, then $du =\frac{1}{t}dt$ and hence $u$ changes from $\ln e=1$ to $\ln e^2=2$; thus we have: $$\int_{e}^{e^2} \frac{1}{t\ln t}dt=\int_{1}^{2} \frac{du }{u}=\ln u|_{1}^{2}\\ =\ln 2-\ln 1=\ln2.$$ where used the properties $\ln e=1$ and $\ln B^x=x\ln B$.